Wanted: Simple integration theoryIf the function $f$ is bounded and continuous a.e. on $[a,b]$, then $f$ is Riemann integrable, and any such sequence of sums over equally-spaced partitions will have to converge to the Riemann integral.

How to show $\Im\{z\cot(z)\}$ is not $0$ in the first quadrant?@DavidC.Ullrich : Maybe this is the only reasonable way to prove such a thing, but it sure seems strange. You get a nice representation using Herglotz' theorem that results in a classical sum expansion of $\cot$ from just knowing the residues. You can, of course get it from the sum expansion, but that's worse than the indirect method I mentioned above!

“Every function can be represented as a Fourier series”?If you're going to sample the signal over an interval $T$ of time, then the Fourier series with exponentials $e^{i2n\pi/T}$ for $n=0,\pm 1,\pm 2,\cdots$ will represent the signal over $[0,T]$. Obviously outside of that interval $[0,T]$, the Fourier series extends periodically, even though the signal may do something else.